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Zero-scoring cribbage hand

I am an avid cribbage player and have enjoyed the game for close to fifty years. I've never had a perfect twenty-nine hand but I've had three twenty-eight hands. I know that the odds of getting the twenty-nine hand are approximately one in a little over three million [ Actually, it's one in about 200,000 - Ed ]. Recently something very strange occurred to me. I dealt the first hand of a new game and here is how the scoring went:

I have never heard of anyone else accomplish this feat. As you are aware the dealer of a hand will always score at least one point on a "go" or a "last card". So I am wondering - what are the odds of such an event occurring? I'm pretty decent with math problems however I'm not sure I even know exactly what equations are required to calculate the odds of getting a perfect "imperfect" hand. Can you help me figure out the end result?

To work out the odds of a zero-scoring cribbage hand (apart from the mandatory one point for go), we need to calculate how many such hands there are, and then divide that into the number of all possible cribbage hands. It is a tricky problem because we need to make no points in the play, assuming correct play. Anyone care to tackle it?

Without considering one of the cards in your hand, if you have three sixes in your hand and a six is turned up, how many points is that? People I play with are insistent it is only 12 points. However, I think it counts three runs of six which would make it 18 points. Do you know the correct answer?

You cant have three runs of six as you would be using the same pair twice, any 4 cards of the same face value score 12, with the exception of 5's as they have the 15 factor added for 8 + 12 = 20.

6D, 6H, 6S, 6C, SO 6D+6H=2, 6D+6S=2, 6D+6C=2 -6PTS-- 6D is now redundant as has been used with other 3

6H, 6S, 6C, are left, SO NOW 6H+6S=2, 6H+6C=2 -4PTS- 6H is now redundant as has been used with other 3, was used with 6D above

6S AND 6C are now the only two cards not used together, they are the final pair for 2 pts, add in the 6 original points plus the 4 from above gives a total of 12, all cards have now been paired and cannot be used together again. Try this with any 4 cards of the same face value, place them down and take one card pair with the others and remove when you used it once with all other cards.

If you want a real challenge, try holding 4 sixes and assuming a nine turns up, then assume a 3 turns up, these are two hands where people miss the points

You do not have "Runs of 6's" you have pairs of sixes. Since 4-of-a-kind is really just 6 sets of pairs, score each pair worth two points. It is really as simple as that. Ignore the numerical value of the cards for a second because it applies to all cards when you are counting pairs. Look at the suits and then count your pairs. Clubs-Heart (1) Clubs-Spades (2) Clubs-Diamonds (3) Hearts-Spades (4) Hearts-Diamonds (5) Spades-Diamonds (6). That is all the possible pairs you can make without duplicating or repeating yourself. Since there are six pairs and each pair is worth two points all 4-of-a-kind are worth 6x2=12 points.